ORCA: CIS calculation for Helium atom

Question

I want to understand ORCA output for CIS calculations. From the input file

! RHF 6-31G
%cis
    Nroots 1
    MaxDim 2
end
    
%coords
    CTyp xyz       # the type of coordinates = xyz or internal
    Charge 0       # the total charge of the molecule
    Mult 1         # the multiplicity = 2S+1
    Units Angs     # the unit of length = angs or bohrs
    coords
        He        0.000000       0.00000        0.00000
    end
end
%output
    Print[ P_Basis ] 2
    Print[ P_MOs ] 1
end 

I get SCF results

----------------
ORBITAL ENERGIES
----------------

  NO   OCC          E(Eh)            E(eV) 
   0   2.0000      -0.914127       -24.8747 
   1   0.0000       1.399859        38.0921 
------------------
MOLECULAR ORBITALS
------------------
                      0         1   
                  -0.91413   1.39986
                   2.00000   0.00000
                  --------  --------
  0He  1s        -0.592081  1.149818
  0He  2s        -0.513586 -1.186959

and CIS results

-----------------------------
CIS-EXCITED STATES (SINGLETS)
-----------------------------

the weight of the individual excitations are printed if larger than 1.0e-02

STATE  1:  E=   1.911194 au     52.006 eV   419458.5 cm**-1 <S**2> =   0.000000
     0a ->   1a  :     1.000000 (c=  1.00000000)
-----------------------
CIS/TD-DFT TOTAL ENERGY
-----------------------

    E(SCF)  =     -2.855160426 Eh
    DE(CIS) =      1.911193623 Eh (Root  1)
    ----------------------------- ---------
    E(tot)  =     -0.943966803 Eh


-------------------------   --------------------
FINAL SINGLE POINT ENERGY        -0.943966803492
-------------------------   --------------------

The CIS method states

1. The solution of the Hartree-Fock equations gives a set of $ M $ spin orbitals, and to construct Slater's determinant use only $ N $, which orrespond to the minimum orbital energies (Aufbau principle).

2. Some of the remaining $ M-N $ functions (corresponding to virtual orbitals) are used to construct additional Slater determinants.

These determinants are obtained by substituting a certain number of spin orbitals of the original determinant $ \Phi^{0} $ for the corresponding number of virtual spin orbitals. The obtained determinants are called excited and are denoted as $\Phi_K $.

The wave function has the form: \begin{equation*}\label{} \Phi = \Phi^{0} + \sum_{K = 1}^{L} C_K\Phi_K. \end{equation*}

3. The search for $ \Phi $ is reduced to the variational problem of minimizing electronic energy by varying the coefficients $ C_K $.

4. When forming the set $ \{\Phi_K\}_ {1, \ldots, L} $ often are limited to determinants, single- and double-excited with respect to $ \Phi^{0} $.

\begin{multline*}\label{ECIS}\tag{ECIS} E_{CI} = \langle{\Phi^{0} + \sum_{K = 1}^{L} C_K\Phi_K}{|\hat{H}|}{\Phi^{0} + \sum_{P = 1}^{L} C_P\Phi_P}\rangle = \\ = \langle{\Phi^{0}}{|\hat{H}|}{\Phi^{0}\rangle} + \langle{\sum_{K = 1}^{L} C_K\Phi_K}{|\hat{H}|}{\sum_{P = 1}^{L} C_P\Phi_P}\rangle \\ \approx E_{HF} + \Delta E. \end{multline*}

For $L = 2$, for example, from \eqref{ECIS} $$ E_{CI} = E_{HF} + |C_1|^2 H_{11} + |C_2|^2 H_{22} + 2C_1C_2 H_{12} $$ So, where $C_K$'s in ORCA output and how did it calculate the excitation energy (where to read the sane information)? Globally, what do all the obtained values mean and how can they be compared with the theory?

Answer 1

What you have obtained in this calculation is the energy of the first excited state according to CIS. In practice, the CI expansion coefficients are obtained as the eigenvectors of the Hamiltonian matrix. The corresponding eigenvalues give the CI electronic energies.

For the 6-31G basis set, you have two AOs, hence two MOs. This allows you to have two configuration state functions in CIS: $\Psi_0$ and $^1{\Psi_1^2}$.

The Hamiltonian matrix is defined by $$\left[\begin{array}{cc} \langle\Psi_0|\hat{H}|\Psi_0 \rangle & \langle\Psi_0|\hat{H}|^1{\Psi_1^2} \rangle \\ \langle ^1{\Psi_1^2} | \hat{H}|\Psi_0 \rangle & \langle ^1{\Psi_1^2} | \hat{H}|^1{\Psi_1^2} \rangle \\ \end{array}\right] = \left[\begin{array}{cc} H_{11} & H_{12} \\ H_{21} & H_{22} \end{array}\right] $$

The matrix element $\langle\Psi_0|\hat{H}|\Psi_0 \rangle =H_{11}$ is the Hartree-Fock energy, $E_{\rm HF}$ (which is E(SCF) = -2.855160426 Eh in your calculation). The above matrix is usually presented by subtracting this energy from the diagonal elements.

$$\left[\begin{array}{cc} H_{11} - E_{\rm HF} & H_{12} \\ H_{21} & H_{22}- E_{\rm HF} \end{array}\right] = \left[\begin{array}{cc} 0 & H_{12} \\ H_{21} & H_{22}- E_{\rm HF} \end{array}\right] $$

CIS of He atom with 6-31 G basis set is actually not a great choice because $H_{12}$ (and $H_{21}$, since the Hamiltonian matrix is Hermitian) vanishes according to Brillouin's theorem.

So, we have a $2\times 2$ diagonal matrix, where the element $H_{22}-E_{\rm HF}$ has the value DE(CIS) = 1.911193623 Eh (Root 1) according to your Orca calculation.

The first excited state's energy is $H_{22}=E_{\rm HF}+1.911193923$, which is E(tot) = -0.943966803 Eh in your calculation.

A somewhat non-trivial example where you will not have a diagonal matrix, hence you can learn about correlation energy correction to the ground state, could be He atom's CISD with the same basis set.